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telo118 [61]
3 years ago
5

What is measurement​

Physics
2 answers:
Anna [14]3 years ago
6 0

This is the assignment of a number to a characteristic of an object or event, which can be compared with other objects or events.

klasskru [66]3 years ago
4 0
Measurement is the act of determining a target size, length, weight, capacity or other aspect
You might be interested in
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
2 years ago
Read 2 more answers
A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the
Mariana [72]

The difference in the pressure between the inside and outside will be 369.36 N/m²

<h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

\rm  \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\  P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\  \triangle p=369.36 \ N/m^2

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ1

3 0
2 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Một bóng đèn có hiệu điện thế định mức là 110V. Đặt vào hai đầu bóng đèn các hiệu điện thế sau đây, hỏi trường hợp nào dây tóc c
Leokris [45]

Answer:

fhaiisjrbekxkrkebxnsiiwj4bnds

8 0
3 years ago
A concrete column has a diameter of 380 mm and a length of 2.6 m . if the density (mass/volume) of concrete is 2.45 mg/m3, deter
grigory [225]

The volume of the column is

(π) · (r²) · (length) =

(π) · (0.19 meter)² · (2.6 meters) =

(π) · (0.036 m²) · (2.6 m) =

0.294 m³ .

The density is 2,450 kg/m³ (VERY very dense, heavy concrete)

so the weight of the column is (mass)·(gravity) or

(density) · (volume) · (gravity) =

(2,450 kg/m³) · (0.294 m³) · (9.81 m/s²) =

(2,450 · 0.294 · 9.81) (kg · m³· m) / (m³ · s²) =

7,066 kg-m/s² = 7,066 Newtons .

But 9.81 Newtons = 2.20462 pounds on Earth (the weight of 1 kilogram of mass), so we have

(7,066 N) · (2.205 pound/9.81 N) =

(7,066 · 2.205 / 9.81) pounds =

1,588 pounds .

5 0
3 years ago
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